The K_{sp} values of Ag_2CrO_4, AgCl, AgBr, a | vedclass

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(A) To determine the order of precipitation,we calculate the concentration of $Ag^+$ required for each salt to start precipitating. For $AgCl, AgBr, A

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$Ag_2CrO_4$

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(A) To determine the order of precipitation,we calculate the concentration of $Ag^+$ required for each salt to start precipitating. For $AgCl, AgBr, AgI$ ($1$:$1$ salts),$[Ag^+] = K_{sp} / [X^-]$. For $Ag_2CrO_4$ ($1$:$2$ salt),$[Ag^+] = \sqrt{K_{sp} / [CrO_4^{2-}]}$. Assuming equal concentrations of anions $(C)$,the required $[Ag^+]$ values are: $[Ag^+]_{AgCl} = 1.8 	imes 10^{-10} / C$ $[Ag^+]_{AgBr} = 5.0 	imes 10^{-13} / C$ $[Ag^+]_{AgI} = 8.3 	imes 10^{-17} / C$ $[Ag^+]_{Ag_2CrO_4} = \sqrt{1.1 	imes 10^{-12} / C} = 1.05 	imes 10^{-6} / \sqrt{C}$. Since $C$ is small,the $[Ag^+]$ required for $Ag_2CrO_4$ is significantly higher than the others,meaning it will precipitate last.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

The $K_{sp}$ values of $Ag_2CrO_4, AgCl, AgBr,$ and $AgI$ are $1.1 \times 10^{-12}, 1.8 \times 10^{-10}, 5.0 \times 10^{-13},$ and $8.3 \times 10^{-17}$ respectively. If $AgNO_3$ solution is added to a solution containing equal moles of $NaCl, NaBr, NaI,$ and $Na_2CrO_4$,which will precipitate last?

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